Solve the following definite integral:

x goes from 0 to 2a

x³ (2ax - x³)^1/2 dx

Where  2F1(a,b;c;x) is hyper geometric function......................this integeration is not in course..you should not stick your mind on such question........In JEE and other Exam....you got only those questions which are in course and which can be easily solved by you................

This question is submitted by you guys twice in the forum....so plzz...

No actually its a definite integral nd not indefinite. I have seen such questions its in jee syllabus.

 can neone give the soln...i guess its vry much in syllab

I think the answer can be obained quite easily, if I'm not too mistaken....Take x^(1/2) common from (2ax-x^3)^(1/2)...the whole expression becomes : int  x^(7/2) (2a - x^2)^(1/2) ...then take x=sqrt(2a)*sin(t)..The final form is in sin^m t cos^m t, which can be easily evaluated...and the limits put.....Please post your comments....& help me out with my rates if this method is correct..

Very nice approach boss............................

I forgot this identity...Lots of Hattt to u !!!!!! :D 

 23.     

   Case (1)

: If m is odd and n is either even or odd

Case (2)

: If m is even and n is even

Case (3)

: If m is even and n is odd.

Use these formulaes................i gues nilesh is correct

Those are Walli's formulae and they can be derived from gamma function...thnx!